Keita left camp three days ago on a journey into the jungle. The three days of his journey can be described by displacement (distance and direction) vectors ${\vec{d_1}}$, ${\vec{d_2}}$, and ${\vec{d_3}}$. ${\vec{d_1}} = (7,8)$ ${\vec{d_2}} = (6,2)$ ${\vec{d_3}} = (2,9)$ (Distances are given in kilometers, $\text{km}$.) How far is Keita from camp at the end of day three?
Solution: We can find Keita's displacement, $\vec d$, from camp at the end of day three by summing the daily displacement vectors. $\vec d = {\vec{d_1}} + {\vec{d_2}} + {\vec{d_3}} = {(7,8)} + {(6,2)} + {(2,9)} = (15, 19)$ We can find the magnitude of $\vec d$ using the Pythagorean theorem, which will tell us how far Keita is from camp at the end of day three. $\begin{aligned} \|\vec d \|^2 &= 15^2 + 19^2\\\\ \| \vec d \| &= \sqrt{225 + 361}\\\\ \| \vec d \| &= \sqrt{586}\\\\ \| \vec d \| &\approx 24.2 \text{ km} \end{aligned}$ Finding the direction of $\vec d$ will tell us what direction Keita is from camp at the end of day three. $\vec d$ is pointing in the first quadrant with an $x$ -component of $15$ and a $y$ -component of $19$. We can find the direction of any vector in the first quadrant using the arctangent function. $\begin{aligned} \tan \theta &= \dfrac{y}{x}\\\\ \tan \theta &= \dfrac{19}{15}\\\\ \theta &= \arctan{ \left ( \dfrac{19}{15} \right ) } \\\\ \theta&\approx 52^\circ \end{aligned}$ Keita is $24.2$ kilometers from camp at the end of day three. Keita is $52^\circ$ from camp at the end of day three.